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\def\lf{\ \hfil\break}       % Neue Zeile ohne Einr"ucken, 'linefeed'
\def\Lf{\vskip0.8mm\ni}
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\cl{\bf On Curves Given By Their Support Function}
\vskip3mm
     
\ni
This note is about smooth, closed, convex curves in the plane
and how to define them in terms of their so-called
{\it Minkowski support function $h$\/}. For quick reference we
first show how, in 3D-XplorMath, $h$ can be modified by specifying
parameters. Then we begin with a more 
general class of geometric objects, namely {\it convex bodies\/}.
\vskip2mm
\cl{1. Parameter Dependent Formulas}
\ni
In 3D-XplorMath, the support function $h$ is given in terms of Fourier 
summands:
$$ \eqalign{
h(\varphi) :=\ \    &aa + bb\cos(\varphi) + cc\cos(2\varphi) +     \cr
                                 & dd\cos(3\varphi) + ee\cos(4\varphi) + f\!f\cos(5\varphi).
 }$$
\noindent
In terms of this function we define the following curve:
$$ \eqalign{
c(\varphi) :=    
h(\varphi)\cdot\left(\matrix{\cos(\varphi) \cr \sin(\varphi)}\right) +
         h'(\varphi)\cdot\left(\matrix{-\sin(\varphi) \cr +\cos(\varphi)}\right). 
}$$
\noindent
Differentiation shows that $c$ is given in terms of its unit normal and tangent
vectors and the function $h$:
$$ \eqalign{
c'(\varphi) =  
    (h+h'')(\varphi)\cdot\left(\matrix{-\sin(\varphi) \cr +\cos(\varphi)}\right). 
}$$
\noindent
One obtains curves with nonsingular parametrization ($|c'|>0$) if $aa$ is
choosen large enough. And since $h(\varphi)$ equals the scalar product
between $c(\varphi)$ and the unit normal $n(\varphi)=(\cos(\varphi), \sin(\varphi))$
one has a simple geometric interpretation: $h(\varphi)$ is the distance of the
tangent at $c(\varphi)$ from the origin.

\vskip2mm
\cl{2. Background And Explanations}
\ni
A convex body in $\Bbb R^n$ is a compact subset $\Bbb B$ having 
non-empty interior and such that it includes the line segment joining 
any two of its points. A hyperplane $H$ in $\Bbb R^n$ is called a 
supporting hyperplane of $\Bbb B$ if it contains a point of $\Bbb B$
and if $\Bbb B$ is included in one of the two halfspaces defined by $H$.
It is not difficult to show that every boundary point of $\Bbb B$ lies 
on at least one supporting hyperplane, and that $\Bbb B$ is the 
intersection of all such halfspaces. 
\Lf
A smooth, closed, planar curves $c$ is called {\it convex\/} if its 
tangent at each point intersects $c$ only at that one point. 
The complement in $\Bbb R^2$ of such a curve has a single
bounded component, the {\it interior\/} of the curve, and one 
unbounded component, its {\it exterior\/}.  The curve is the 
boundary of its interior, and we denote by $\Bbb B$ the curve 
together with its interior. It is easy to see that $\Bbb B$ is a
convex body  in $\Bbb R^2$, as defined above, and in fact 
the tangent line at any point of $c$ is the unique supporting
hyperplane (= line!) containing that point. (There are of course 
more general planar convex bodies. For example if $P$ is a 
closed polygon in $\Bbb R^2$ together with its interior, then $P$
is a convex body, but there are infinitely many supporting 
lines through each vertex, while the supporting line containing 
an edge contains infinitely many points.)
\Lf
Now let $O$ be some interior point of $c$ and take $O$ as
the origin of a cartesian coordinates by fixing a ray from $O$ as 
the positive $x$-axis. With respect to these coordinates, at each 
point $p$ on $c$ the outward directed unit normal at $p$ will have the 
form $n(\varphi) = (\cos(\varphi),\ \sin(\varphi))$
where $\varphi = \varphi(p)$ satisfies $0 \le \phi \le 2\pi$.
If we as usual think of $\Bbb S^1$ as the interval 
$[0,2 \pi]$ with endpoints identified, then it can be shown that 
the map $p \mapsto \varphi(p)$ is a smooth one-to-one map of 
$c$ with $\Bbb S^1$, so that the inverse map gives a 
parametrization $c(\varphi)$ of the curve by $\Bbb S^1$. 
(This just says that given any direction in the plane, 
there is a unique point $p$ on $c$ where the outward normal 
has that direction, and the point $p$ varies smoothly with the direction.)
\Lf
The Minkowski {\it support function\/} for the curve $c$ is the function
$h$ defined on $\Bbb S^1$ by letting $h(\varphi)$ be the 
distance from the origin of the line of support (or tangent) through
$c(\varphi)$, that is  $h(\varphi) := n(\varphi) \cdot  c(\varphi)$,
the scalar product of $c(\varphi)$ and $n(\varphi)$. From this definition 
it is easy to reconstruct the curve in terms of its support function as in part 1.

\vskip2mm
\cl{3. Things To Observe}
\ni
Recall one has in any parametrization the curvature formula \lf
\centerline{$ n'(t) = \kappa(t)c'(t)$,}
which in the present case reduces to: \lf
\centerline{$ 1/ \kappa(\varphi) = h(\varphi) + h''(\varphi) =|c'(\varphi)|.$}
\Lf
Clearly $aa$ has to be large enough to make $\kappa$ positive 
and the parametrization nonsingular. Adding a linear combination of 
$\cos(\varphi)$ and $\sin(\varphi)$ to the support function 
corresponds to a change of only the origin, the shape of the curve stays the
same. The $bb\cos(\varphi)$-term in the support function is therefore not really necessary, but one can use it to see how the parametrization of the curve 
changes.
\Lf
The $\cos$-terms in even multiples of $\varphi$ make up the even part 
$(h(\varphi)+h(\varphi+\pi))/2$ of $h$. The origin is the {\it midpoint} of curves with even support function. If $h$ is odd except for the constant term, i.e., \lf
\centerline{$h(\varphi) = aa + (h(\varphi) - h(\varphi+\pi))/2$,} 
then one obtains curves of constant width $w$ where: \lf
\centerline{ {\it w} $ = h(\varphi) + h(\varphi+\pi)= 2\,aa$.}
The default curve in 3D-XplorMath is such a curve of constant width and the default morph
shows a family of such curves. We emphasize the width of our curves by drawing them together 
with their pairs of parallel tangents. Since the (non-)constancy of the distance between these
parallel tangents is difficult to see we have added a circle of the same width (= diameter).
One cannot easily recognize how many extrema the curvature 
$\kappa(\varphi)$ has. To see it clearly we
recommend selecting  the entry {\it Show Osculating Circles}
from the Action Menu, since the evolute has a cusp at every 
extremal value of $\kappa$.



\bye